Ta có: \(3^{n+2}-2^{n+2}+3^n-2^n\)
\(=\left(3^n.3^2+3^n\right)-\left(2^{n-1}.2^3+2^{n-1}.2\right)\)
\(=\left[3^n.\left(3^2+1\right)\right]-\left[2^{n-1}.\left(2^3+2\right)\right]\)
\(=3^n.10-2^{n-1}.10\)
\(=\left(3^n-2^{n-1}\right).10⋮10\)
\(\Rightarrow3^{n+2}-2^{n+2}+3^n-2^n⋮10\left(đpcm\right)\)
Ta có:3n+2-2n+2+3n-2n
=(3n+2+3n)-(2n+2+2n)
=3n.(32+1)-2n.(22+1)
=3n.10-2n.5
=3n.10-2n-1.2.5
=3n.10-2n-1.10
=10.(3n-2n-1) \(⋮\)10
=> đpcm
Ta có : 3n+2 - 2n+2 + 3n - 2n
= 3n+2 + 3n - 2n+2 - 2n
= ( 3n . 32 + 3n ) - ( 22 . 2n + 2n )
= 3n . ( 9 + 1 ) - 2n . ( 4 + 1 )
= 3n . 10 - 2n . 5
= 10 . 3n - 10 . 2n-1
= 10 . ( 3n - 2n-1 )
Ta thấy 10 \(⋮\) 10 => 10 . ( 3n - 2n-1 ) \(⋮\) 10
Vậy 3n+2 - 2n+2 + 3n - 2n \(⋮\) 10
Chúc bn hok tốt !
